Bài 2: Phương trình bậc nhất một ẩn và cách giải

TH1: m=0

Phương trình sẽ là -6=0(loại)

TH2: m<>0

=>PT sẽ có nghiệm duy nhất là 6/m^2

\(\Leftrightarrow\left(4-3x\right)\left(4+3x\right)=0\)

=>x=4/3 hoặc x=-4/3

`16-9x^2 = 0`

`9x^2=16`

`x^2=16 : 9`

`x^2 =16/9`

`=>` \(x=\sqrt{\dfrac{16}{9}}\)

`x=4/3` hoặc `x=-4/3`

Vậy `S = {4/3; -4/3}`

9x² = 16

x² = 16/9

x² = (4/3)²

x = \(\pm\) 4/3

=> x ∈ {4/3 ; -4/3}

`-> x^2 - 2x - 2 = x^2 + 18x - 40`

`<=> 38 = 20x`

`=> x = 19/10 = 1,9`

1.

đkxđ : x khác 5 ; x khác 2 

pttđ :

\(\dfrac{\left(2+x\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(2-x\right)\left(x-5\right)}{\left(2-x\right)\left(x-5\right)}-\dfrac{6\left(x-5\right)}{\left(2-x\right)\left(x-5\right)}=0\)

<=> \(\dfrac{2^2-x^2}{....}+\dfrac{\left(6-3x\right)\left(x-5\right)}{....}-\dfrac{6x-30}{...}=0\)

<=> \(\dfrac{4-x^2}{...}+\dfrac{6x-30-3x^2+15x}{...}-\dfrac{6x-30}{...}=0\)

<=> \(4-x^2+6x-30-3x^2+15x-6x+30=0\)

<=> \(-4x^2+15x+4=0\)

\(\Delta=\left(15\right)^2-4.-4.4=289>0\)

=> pt có 2 nghiệm pb

\(x_1=\dfrac{-15+\sqrt{289}}{-8};x_2=\dfrac{-15-\sqrt{289}}{-8}\)

<=> \(x_1=-\dfrac{1}{4}\left(tm\right);x_2=4\left(tm\right)\)

<=>(x+2) (2-x)+3(x-5) (2-x)=6(x-5)

2.

\(\Leftrightarrow x^2\left(1-\dfrac{3x^2}{7}\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\1-\dfrac{3x^2}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\\dfrac{3x^2}{7}=1\Rightarrow3x^2=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x^2=\dfrac{7}{3}\Rightarrow x=\sqrt{\dfrac{7}{3}}\end{matrix}\right.\)

\(\Leftrightarrow x^2+4x+4+3>=x^2-9\)

=>4x+7>=-9

=>4x>=-16

hay x>=-4

`(x-3)(x+3) <= (x+2)^2 +3`

`<=> x^2 -9 <=x^2 + 4x +7`

`<=> x^2 -9 -x^2 -4x -7 <= 0`

`<=> -4x-16 <=0`

`<=> -4x <= 16`

`<=> x <= -4`

\(x^2-6x+9< x^2-5x+4\\ x^2-6x+9-x^2+5x-4< 0\\ -x+5< 0\\ -x< -5\\ x>5\)

x² - 6x + 9 < x² - 5x + 4

x² - 6x + 9 - x² + 5x - 4 < 0

- x + 5 < 0

- x < - 5

x > 5 

2x + \(\dfrac{2x+1}{2}\) > 3x - \(\dfrac{1}{5}\)

\(\Leftrightarrow\) 2x.10 + (2x+1).5 > 3x.10 - 1.2

\(\Leftrightarrow\) 20x + 10x + 5 > 30x - 2

\(\Leftrightarrow\) 20x + 10x - 30x > - 2 - 5

\(\Leftrightarrow\) 0x > -7

Vậy: S = \(\varnothing\)

=>20x+10x+5>60x-2

=>30x+5>60x-2

=>-30x>-3

hay x<1/10

Xin lỗi tại tui ko biết việc trình bày nên nếu có chỗ nào không hiểu thì thứ lỗi!!!!!

2x-5/6 + 22-7x/4 > 5-2/3 - x - 5x+2/4

= 2x-5/6 + 22-7x/4 > 1 - x - 5x+2/4

= 2x-5/6.12 + 22-7x/4.12 > 1.12 - x . 12 - 5x+2/4 , 12

= 2(2x-5) + 3 (-7x+22) > 12 -x . 12 - 3 (5x + 2)

= - 17x + 56 > -27x + 6 = -17x+ 56 - 56 > -27x + 6 - 56

= -17x > -27x - 50

= -17x + 27x > - 27x - 50 + 27x

= 10x : 10 > -50 : 10

= x > -5.

\(\dfrac{2x-5}{6}+\dfrac{22-7x}{4}>\dfrac{5-2}{3}-x-\dfrac{5x+2}{4}\\ \Leftrightarrow\dfrac{2\left(2x-5\right)}{6.2}+\dfrac{3\left(22-7x\right)}{4.3}>\dfrac{3.4}{3.4}-\dfrac{12.x}{12}-\dfrac{3\left(5x+2\right)}{3.4}\\ \Leftrightarrow\dfrac{4x-10}{12}+\dfrac{66-21x}{12}>\dfrac{12}{12}-\dfrac{12x}{12}-\dfrac{15x+6}{12}\\ \Leftrightarrow4x-10+66-21x>12-12x-15x-6\\ \Leftrightarrow-17x+56>6-27x\\ \Leftrightarrow-17x+56-6+27x>0\\ \Leftrightarrow10x+50>0\\ \Leftrightarrow10x>-50\\ \Leftrightarrow x>-5\)

Vậy bpt có nghiệm `x > -5`

\(\dfrac{2x-5}{6}+\dfrac{22-7x}{4}>\dfrac{5-2}{3}-x-\dfrac{5x+2}{4}\)
\(\Leftrightarrow\dfrac{2\left(2x-5\right)}{6.2}+\dfrac{3\left(22-7x\right)}{4.3}>\dfrac{4\left(5-2\right)}{3}-\dfrac{12x}{12}-\dfrac{3\left(5x+2\right)}{4.3}\)
\(\Leftrightarrow4x-10+33-29x>20-8-12x-15x-6\)
\(\Leftrightarrow23-25x>6-27x\)
\(\Leftrightarrow27x-25x>6-23\)
\(\Leftrightarrow2x>-17\)
\(\Leftrightarrow x>-8,5\)
Vậy...

\(\dfrac{2x-5}{6}+\dfrac{22-7x}{4}>\dfrac{5-2}{3}-x-\dfrac{5x+2}{4}\\ \Leftrightarrow\dfrac{2\left(2x-5\right)}{6.2}+\dfrac{3\left(22-7x\right)}{4.3}>\dfrac{3.4}{3.4}-\dfrac{12.x}{12}-\dfrac{3\left(5x+2\right)}{3.4}\\ \Leftrightarrow\dfrac{4x-10}{12}+\dfrac{66-21x}{12}>\dfrac{12}{12}-\dfrac{12x}{12}-\dfrac{15x+6}{12}\\ \Leftrightarrow4x-10+66-21x>12-12x-15x-6\\ \Leftrightarrow-17x+56>6-27x\\ \Leftrightarrow-17x+56-6+27x>0\\ \Leftrightarrow10x+50>0\\ \Leftrightarrow10x>-50\\ \Leftrightarrow x>-5\)

Vậy bpt có nghiệm `x> -5`

`[3x+5]/2-1 <= [x+2]/3+x`

`<=>[3(3x+5)-6]/6 <= [2(x+2)+6x]/6`

`<=>9x+15-6 <= 2x+4+6x`

`<=>x <= -5`

Vậy `S={x|x <= -5}`

\(\dfrac{3x+5}{2}-1\le\dfrac{x+2}{3}+x\\ \Leftrightarrow3\left(2x+5\right)-6\le2\left(x+2\right)+6x\\ \Leftrightarrow9x+15-6\le2x+4+6x\\ \Leftrightarrow9x-2x-6x\le4-15+6\\ \Leftrightarrow x\le-5\)

Vậy ...